Question

Note: Base 2= B2
logB2(y+2)-1=logB2(y-2)

Answer 1

@mathmale

Answer 2

shay, I'll be right with you. Hope you've been able to solve that other problem. Have you?

Answer 3

yes and thank you

Answer 4

Shay, I'm back.

Answer 5

logB2(y+2)-1=logB2(y-2)
can be re-written as logB2(y+2)-logB2(y-2)=1.

Please say whether you agree or disagree.

Answer 6

yes, i agree

Answer 7

don't you have to divide now?

Answer 8

Use Rule #2 that I shared with you earliler:

earlier, sorry

Rule #2: log a-log b =log (a/b).

can you now apply Rule #2 to combine the 2 terms on th left side of your equation?

Answer 9

so it would be:
logB2 y+2/y-2=1 ?

Answer 10

That's pretty much correct, Shay, but it's important that you enclose y+2 and y-2 inside parentheses. Would you plese do that next time.

But for now,


write the equation 2 = 2

Answer 11

and as the exponent of the first " 2 " please write logB2 [(y+2)/(y-2)].
(see how I used parentheses?

Answer 12

and for the exponent of the 2nd " 2 " use 1.

this new equation reduces to what?

Note that we're using the inverse function property here, to simplify things.

Answer 13

where did you get the 2 from? the one that's on the right side of the equal sign

Answer 14

Yes, that's right. Perhaps you have a different way of doing this. if you find this confusing, I'll write everything out.

Answer 15

so would it be
2^1= (y+2)(y-2)

Answer 16

Here's what I'd expect to get as a result:\[2=\frac{ y+2 }{y-2 }\]

Answer 17

would you multiply next or um, idk?

Answer 18

the main difference is that y ou multiplied the ( )( ), whereas, I divided: ( ) / ( )

OK with this?

then all you have to do is to solve the equation above for y.

I'll be back in a few minutes in case you need to work on this further.
so nice working with you!

Answer 19

no i dont know what to do

Answer 20

OK. If ...
\[2 = \frac{ y+2 }{ y-2 }\] and we want to solve for y, one of the e3asier methods is to multiply both sides of the equation by the LCD, which is y-2.

Then \[2(y-2) = y+2.\]

Multiply out the left side. What do you get?

Answer 21

I get 2y-4=y+2.

How would you solve this equation for y?

Answer 22

Hint #1: subtract y from both sides of the equation.
Hint #2: add 4 to both sides of the equation.
Solve for y.

y = ?

Answer 23

BRB

Answer 24

Shay,

If 2y-4=y+2,
then subtracting y from both sides gives us y-4=2;
adding 4 to both sides gives us y=2+4, or y=6. that's the solution.

Please review this problem and determine whether or not you'd like to discuss any part of the solution in greater depth.

BRB

Answer 25

why did you out 2(y-2)=y+2 like how did you do that?

Answer 26

Shay, I'm trying to solve the equation \[2=\frac{ y+2 }{ y-2 }\] for y.

To do this, I multiply both sides of the equation by the denominator y-2, to eliminate the fraction:

2(y-2)=y+2.

Are you OK with that?

Answer 27

After multiplying out the first term, I get
2y-4=y+2

and then 2y-y-4=2
and then y - 4 = 2, and, finally,
y=6.

Answer 28

Shay, please note that if you substitute y=6 into \[2=\frac{ y+2 }{ y-2 }\]the resulting equation is true! This is the "solution" to the problem in logB2 that you presented.

Answer 29

Hope you'll reply. If you do, I'll make sure we take this problem all the way to a satisfactory solution.